Two pumps are being considered for purchase for a service life of 10 years. Interest rate...
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Two pumps are being considered for purchase for a service life of 10 years. Interest rate is 5% Alternatives Pump 1 Pump 2 $60,000 $75,000 Apply the present worth technique, which pump should be considered? k Solution: 1. Neither input nor output is the same, calculate the net present worth (NPW) 2. NPW₁ o Initial cost: PW • Equivalent PW for salvage and replacement together. PW= k o Thus NPW₁= Initial Cost Estimated salvage value at end of useful life k Cash flow at the end of 10 years, PW= k. 3. NPW₂ • Initial cost: PW= Useful Life 6 years 12 years Estimated market value, end of 10-year $12,000 $15,000 k • Equivalent PW for salvage and replacement together: PW= k • Thus NPW₂= k Cash flow at the end of 10 years, PW= k. 4. Hence use $10,000 $12,000 Two pumps are being considered for purchase for a service life of 10 years. Interest rate is 5% Alternatives Pump 1 Pump 2 $60,000 $75,000 Apply the present worth technique, which pump should be considered? k Solution: 1. Neither input nor output is the same, calculate the net present worth (NPW) 2. NPW₁ o Initial cost: PW • Equivalent PW for salvage and replacement together. PW= k o Thus NPW₁= Initial Cost Estimated salvage value at end of useful life k Cash flow at the end of 10 years, PW= k. 3. NPW₂ • Initial cost: PW= Useful Life 6 years 12 years Estimated market value, end of 10-year $12,000 $15,000 k • Equivalent PW for salvage and replacement together: PW= k • Thus NPW₂= k Cash flow at the end of 10 years, PW= k. 4. Hence use $10,000 $12,000
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